Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 10.4 due 08/29/2021 at 11:59pm MST
Problem 1. (1 point)
Find the cross product a×bwhere a=h−2,−3,−5iand
b=h5,3,0i.
a×b=h, , i
Solution:
SOLUTION
a×b=
i j k
−2−3−5
530
=
−3−5
3 0
i−
−2−5
5 0
j+
−2−3
5 3
k
= (0+15)i−(0+25)j+ (−6+15)k
=15i−25j+9k
=h15,−25,9i
Answer(s) submitted:
•15
•-25
•9
(correct)
Correct Answers:
•15
•-25
•9
Problem 2. (1 point)
Find the cross product a×bwhere a=h0,5,−2iand b=
h5,−1,−1i.
a×b=h, , i
Find the cross product c×dwhere c=−4i−1j−4kand d=
−1i−5j+4k.
c×d=i+j+k
Solution:
SOLUTION:
a×b=
i j k
0 5 −2
5−1−1
=
5−2
−1−1
i−
0−2
5−1
j+
0 5
5−1
k
= (−5−2)i−(0+10)j+ (0−25)k
=−7i−10j−25k
=h−7,−10,−25i
c×d=
i j k
−4−1−4
−1−5 4
=
−1−4
−5 4
i−
−4−4
−1 4
j+
−4−1
−1−5
k
= (−4−20)i−(−16 −4)j+ (20 −1)k
=−24i+20j+19k
Answer(s) submitted:
•-7
•-10
•-25
•-24
•20
•19
(correct)
Correct Answers:
•-7
•-10
•-25
•-24
•20
•19
1
Problem 3. (1 point)
You are looking down at a map. A vector uwith |u|= 1 points
north and a vector vwith |v|= 10 points northeast. The crossprod-
uct u×vpoints:
A) south
B) northwest
C) up
D) down
Please enter the letter of the correct answer:
The magnitude |u×v|=
Solution:
SOLUTION
By the right hand rule, the cross product points down.
|u×v|=|u||v|sinθ=1(10)sin(45o) = 5√2
Answer(s) submitted:
•d
•7.0710
(correct)
Correct Answers:
•D
•(1*10*sqrt(2) )/2
Problem 4. (1 point)
Think of the letter Xas four vectors starting from the center and
pointing outward. Label the four vectors starting from the top left
and proceeding clockwise as u,v,w,z.
Does u×vpoint in or out of the page? (in/out)
Does u×zpoint in or out of the page? (in/out)
Compute u×w
h, , i
Solution:
SOLUTION:
By the right hand rule, u×vpoints into the page.
By the right hand rule, u×zpoints out of the page.
Since uand ware parallel, their cross product is the zero vector.
Answer(s) submitted:
•in
•out
•0
•0
•0
(correct)
Correct Answers:
•IN
•OUT
•0
•0
•0
2
Problem 5. (1 point)
Find two unit vectors orthogonal to a=h1,−4,−3iand b=
h−3,3,−1i
Enter your answer so that the first non-zero coordinate of the first
vector is positive.
First Vector: h, , i
Second Vector: h,,i
Solution:
SOLUTION:
The cross product of two vectors is orthogonal to both vectors. So
we calculate
a×b=
i j k
1−4−3
−3 3 −1
=
−4−3
3−1
i−
1−3
−3−1
j+
1−4
−3 3
k
= (4+9)i−(−1−9)j+ (3−12)k
=13i+10j−9k
=h13,10,−9i
The magnitude of the cross product is
|a×b|=p(13)2+ (10)2+ (−9)2=√350,
so two unit vectors orthogonal to both are
±a×b
|a×b|=±13
√350 ,10
√350 ,−9
√350
, that is,
13
√350 ,10
√350 ,−9
√350 and −13
√350 ,−10
√350 ,+9
√350
Answer(s) submitted:
•((13)/(sqrt(278)))
•((10)/(sqrt(278)))
•(3/(sqrt(278)))
•
•
•
(incorrect)
Correct Answers:
•0.694879228972303
•0.534522483824849
•-0.481070235442364
•-0.694879228972303
•-0.534522483824849
•0.481070235442364
Problem 6. (1 point)
Find the area of the triangle with vertices:
Q(0,4,4),R(2,3,2),S(−1,1,1).
Solution:
SOLUTION:
Let a=~
QR =h2,−1,−2iand b=~
QS =h−1,−3,−3i. The area
of the triangle is equal to half the lenght of the cross product of
these two vectors.
The corss product is given by a×b=h−3,8,−7i, and it has mag-
nitude |a×b|=p(−3)2+ (8)2+ (−7)2=√122.
Therefore the area of the triangle is 1
2√122.
Answer(s) submitted:
•(11.04536).5
(correct)
Correct Answers:
•5.52268050859363
Problem 7. (1 point)
Find the area of the parallelogram with vertices:
P(0,0,0),Q(−3,−1,−2),R(−3,−2,0),S(−6,−3,−2).
Solution:
SOLUTION:
Choose any three points from the given four. The area of the par-
allelogram is given by the magnitude of the cross product of two
vectors built from those three points.
For instance, choosing the points P,Qand R, we obtain the vectors
~
PQ =h−3,−1,−2iand ~
PR =h−3,−2,0i.
The cross product of these two vectors is h−4,6,3iwith magni-
tude p(−4)2+ (6)2+ (3)2=√61.
Therefore the area of the parallelogram is √61.
Answer(s) submitted:
•sqrt(61)
(correct)
Correct Answers:
•7.81024967590665
3
Problem 8. (1 point)
Find the distance the point P(−5,−2,−1)is to the line through
the two points
Q(−2,−4,0 ), and R(−4,−6,−3 ).
Solution:
SOLUTION:
The distance between a point and a line is the length of the per-
pendicular from the point to the line, here |~
PS|=d. But referring
to triangle PQS,d=|~
PS|=|~
QP|sinθ=|b|sin θ. But θis the an-
gle between ~
QP =b=h−3,2,−1iand ~
QR =a=h−2,−2,−3i.
Thus by definition of the cross product, sinθ=|a×b|
|a||b|and so
d=|b|sinθ=|b||a×b|
|a||b|=|a×b|
|a|=|h8,7,−10i|
|h−2,−2,−3i| =√213
√17
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•3.5396909137248
Problem 9. (1 point)
Find the volume of the parallelopiped with adjacent edges PQ, PR,
PS where
P(−4,5,−3),Q(−2,8,0),R(−5,4,−4),S(2,3,−1).
Solution:
SOLUTION:
Let a=~
PQ =h2,3,3i,b=~
PR =h−1,−1,−1iand c=~
PS =
h6,−2,2i.
a·(b×c) =
233
−1−1−1
6−2 2
=2
−1−1
−2 2
−3
−1−1
6 2
+3
−1−1
6−2
=4,
so the volume of the parallelepiped is |a·(b×c)|=4 cubic units.
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•4
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